如何在JS脚本中正确回显PHP变量？

Question

I have this script that needs to print and it is within a PHP file as I need to pass it options because I am using the jQuery UI Tabs plugin.

Here is what I have:

<?php 
$collapsible = "true";
$active = "2";

$options = array( 'collapsible' => $collapsible, 'active' => $active );
?>
    <script type="text/javascript">
        jQuery(document).ready(function() {
            jQuery( '.tabs' ).tabs({
                collapsible : <?php echo $options["collapsible"]; ?>,
                active : <?php echo $options["active"]; ?>
            });
        });
    </script>

Ok so everything works however the two options collapsible and active isn't effecting it. But if I bypass the php variables and just hardcode the option settings in for collapsible and active, then it works. So I am not sure why the variables have no effect. I've even tried type casting it with (int) for active and (bool) for collapsible but still no dice.

Thanks for looking.

Answer 1

Rather than adding quotes, run the value through json_encode. This will ensure proper escaping as well:

<script type="text/javascript">
    jQuery(document).ready(function() {
        jQuery( '.tabs' ).tabs({
            collapsible : <?php echo json_encode($options["collapsible"]) ?>,
            active : <?php echo json_encode($options["active"]) ?>'
        });
    });
</script>

It also gives you the added benefit of being able to use literal types as opposed to all strings in your PHP:

<?php
$collapsible = true;
$active = 2;

And, per axel.michel suggestion in comments, could be simplified to:

<?php
$options = array('collapsible' => true, 'active' => 2);
?>

<script type="text/javascript">
    jQuery(document).ready(function() {
        jQuery( '.tabs' ).tabs(<?php echo json_encode($options); ?>);
    });
</script>